~~Mathematics Revising Game~~ (1 Viewer)

[Graceofgod]

Re: 回复: Re: ~~Mathematics Revising Game~~

http://www.boredofstudies.org/course...rial_Paper.pdf

Let's get the thread going, shall we?

Try this awesome q 10 b)

Hint: Don't LOOK AT THE ANSWERS!

I prob shouldn't have looked at the answers. They aren't correct... If @ = pi(t)/90 then when t = 0, @ = 0.
However when t = 0, @ = pi/2 as we have been told. Therefore answers wrong

EDIT: Hence the hint? Or am I just wrong?

EDIT: NEVERMIND

 

[Graceofgod]

Re: 回复: Re: ~~Mathematics Revising Game~~

Ok got it:

10bi) d@/dt = pi/90 (2 degrees)
@ = pi(t)/90 + C
@ = pi(t)/90

ii) tan@ = x/10
x = 10tan@
x = 10tan(pi(t)/90)

iii) v = (pi/9)sec^2(pi(t)/90)
v = pi/9 * 1/(cos30)^2
v = 4pi/27 km/s
v = 1600pi/3 km/h

Right, a question: (random catholic trial q10)

A regular pentagon is made into a spinner, labeled A, E, I, O, U. It is equally likely to stop on each one.

i) If the spinner is spun twice, find the prob that it stops on the same letter twice.

ii) how many times must the spinner be spun for it to be 99% sure that it will stop on the letter E once?

 

[Azreil]

Re: 回复: Re: ~~Mathematics Revising Game~~

i) 1 * 0.2 = 0.2
ii)0.99 = 0.2 + 0.2*0.8 + ... + 0.2*0.8^n
Series where a = 0.2, r = 0.8 and n=x
S = a(1-r^n)/(1-r)
=0.2(1-0.8^x)/(1-0.8)
0.99=1-0.8^x
0.8^x=0.01
ln(0.8^x)=ln0.01
x=ln0.01/ln0.8
= 20.637...
= 21

Therefore 21 spins are required to be 99% sure it will stop on the letter E once.

Calculate the area of the region between the curves y = sec^2(x) and y = x between x = 0 and x=pi/4

 

[syriangabsta]

13/52 * 39/51 * 26/50 * 13/49 = 2197/83300 or 0.02637 (4d.p)

Question = find d(sec^2(x))/dx.

SOrry i was just looing thru this thread, and i noticed that Graceofgod has made an error in the probability q?

umm, i dont wanna post any q, but just incase ppl are relying on what they see as thier study, they might wanna know how to really answer the q...

all of it is right except the 1st part,,its not suppsoed to be 13/52, but its supposed to be 52/52 or, 1/1 (coz it doesnt matter what suit u pick 1st)

so, 52/52 x 39/51 x 26/50 x 13/49 = 2197/20825

correct me if im wrong..sorry for disturbing ur thread lol

 

[Graceofgod]

SOrry i was just looing thru this thread, and i noticed that Graceofgod has made an error in the probability q?

umm, i dont wanna post any q, but just incase ppl are relying on what they see as thier study, they might wanna know how to really answer the q...

all of it is right except the 1st part,,its not suppsoed to be 13/52, but its supposed to be 52/52 or, 1/1 (coz it doesnt matter what suit u pick 1st)

so, 52/52 x 39/51 x 26/50 x 13/49 = 2197/20825

correct me if im wrong..sorry for disturbing ur thread lol

haha yep. You're right. lol woops I will go edit it now. -_- *sleeps*

 

[Dota55]

Re: 回复: Re: 回复: Re: ~~Mathematics Revising Game~~

for a)i got 23.0726 --->24 months
doesn't seem right maybe

time for b, i'll assume he starts paying from month 1
i got nth month=PRⁿ(1-Q)ⁿ
where P=$100 000 R=1.008 and Q=0.08

Can anyone confirm this? I got n=24 too, but subbing that in, i think its wrong.

 

[SkimDawg]

Re: 回复: Re: ~~Mathematics Revising Game~~

Calculate the area of the region between the curves y = sec^2(x) and y = x between x = 0 and x=pi/4

I dont think this is possible, i got up to 1/2[tan2pi/4] = 1/2[tan pi/2] = Mathematical error?

 

[ltewierik90]

Re: 回复: Re: ~~Mathematics Revising Game~~

Calculate the area of the region between the curves y = sec^2(x) and y = x between x = 0 and x=pi/4

Integral of sec^2(x) = tan(x)

Area = tan(pi/4) - tan(0)

= 1 - 0
=1 square units
​

For the function y=x^3 - 3x^2 - 9x +1
i. Find the coordinates of any stationary points and determine their nature
ii. Find any points of inflexion

 

[tommykins]

Re: 回复: Re: ~~Mathematics Revising Game~~

I dont think this is possible, i got up to 1/2[tan2pi/4] = 1/2[tan pi/2] = Mathematical error?

the integral of sec²x is tanx.

 

[SkimDawg]

Re: 回复: Re: ~~Mathematics Revising Game~~

the integral of sec²x is tanx.

haha lol my bad, misinterpretation, i thought it was sec^2(2x)

 

[minijumbuk]

Re: 回复: Re: ~~Mathematics Revising Game~~

1=4
2=8
3=24
4=?

Spoiler

1=4, so 4=1

Erm... lim (x->0) sinx/x
lim (x->0) cos x/x
lim (x->0) tan x/x

 

[Azreil]

Re: 回复: Re: ~~Mathematics Revising Game~~

For the function y=x^3 - 3x^2 - 9x +1
i. Find the coordinates of any stationary points and determine their nature
ii. Find any points of inflexion

y'=3x^2-6x-9
y''= 6x-6

TP when y'=0
3x^2-6x-9=0
x^2-2x-3=0
(x-3)(x+1)=0
x= 3, -1

When x=-1,
y''=6(-1)-6 = -12
Hence max t.p.
When x=3
y''=6(3)-6 = 12
Hence min t.p.

(-1,6) max tp, (3, -26) min tp.

Point of inflexion when y''=0
6x-6=0
6(x-1)=0
x=1
CHECK
x 0.5 1 1.5
y'' -3 0 3

(1, -10) point of inflexion

Question 10b 2002 to see diagram:
A circular pizza of radius 20 cm is cut into sectors. Each sector is to be placed on a circular plate that is just large enough to contain that sector.

A sector of pizza is cut where the angle @ at its centre satisfies 0 < @ =< (pi/2)
It is placed on a circular plate, of radius r cm and centre C.

Show that r=10sec(@/2) .
EDIT: Formatting​

 

[lyounamu]

y = ln (1/(cos(x)^1/2)

Find the first derivative.

 

[Aerath]

y' = sin(x^1/2) / 2(x)^1/2cos(x^1/2)

I hope it isn't wrong. =\

 

[ltewierik90]

y' = sin(x^1/2) / 2(x)^1/2cos(x^1/2)

I hope it isn't wrong. =\

y = ln[1/cos(x)^1/2]

= ln [cos(x)^-1/2]
= -1/2 . ln[cos(x)]
= -1/2 . -sin(x)/cos(x)
= sin(x)/2[cos(x)]

​

I think thats the correct answer, i dunno could be wrong

 

[lyounamu]

y = ln[1/cos(x)^1/2]

= ln [cos(x)^-1/2]
= -1/2 . ln[cos(x)]
= -1/2 . -sin(x)/cos(x)
= sin(x)/2[cos(x)]


​

I think thats the correct answer, i dunno could be wrong

Yep, that's right.

Even though you can further simplify it to:

tan(x)/2

 

[Mark576]

y = ln[1/cos(x)^1/2]

= ln [cos(x)^-1/2]
= -1/2 . ln[cos(x)]
= -1/2 . -sin(x)/cos(x)
= sin(x)/2[cos(x)]

​

I think thats the correct answer, i dunno could be wrong

Your working out is incorrect; ln[cos(x)] =/= -sin(x)/cos(x) .

 

[lyounamu]

Your working out is incorrect; ln[cos(x)] =/= -sin(x)/cos(x) .

lol, funny how he got the answer correct though.

I think he just wrote it incorrectly on the board.

 

[Aerath]

Oh - wtf was I doing then....

 

[Mark576]

lol, funny how he got the answer correct though.

I think he just wrote it incorrectly on the board.

Uh, of course? I was just being pedantic :uhhuh: